Thread: How to calculate min acceleration between two points of different velocities

1. Re: How to calculate min acceleration between two points of different velocities

Originally Posted by calltronics
Starting with
$\displaystyle a= \frac {v_c-100}t$

and

$\displaystyle a = \frac {40-v_c} {2-t}$

set them equal, cross-multiply, and manipulate to get $\displaystyle v_c$ by itself:

$\displaystyle \frac {v_c-100} t = \frac {v_c-40}{2-t}\\v_ct - 40 t = (v_c-100)(2-t)\\v_c(2t-2)=-200+40t\\v_c=\frac {200+140t}{2t-2} = \frac {100+70t}{t-1}$

2. Re: How to calculate min acceleration between two points of different velocities

Thank you, it is so easy when you can see it.

3. Re: How to calculate min acceleration between two points of different velocities

Originally Posted by calltronics
The total time is now the variable.
The acceleration is fixed at 0.7m/s^2
Same type of solution Passing through "Point C".
Solved for either t1 or t2 as a minimum total time.
If you set a = 0.7m/s^2 and solve for t1 and t2 then the result is:

t1 = 0.5208 seconds
d1 = 0.147m
v_c = 0.464 m/s
t2 = 0.6065 seconds
d2 = 0.153m
T_total = 1.127 second

In reviewing my previous posts with the solution to your first question I see that I made a typo, which I have corrected in bold font below:

Originally Posted by ebaines
Apply the quadratic equation to solve, and you find that t = 0.818665 seconds. This yields a = 0.165 m/s^2, point c at 0.137m, and v_c = 0.235m/s.

4. Re: How to calculate min acceleration between two points of different velocities

wow glad you spotted that I am going to motion trials tomorrow.
Bit of difference between 0.818 m/s^2 and 0.818s
Thing is without thinking I used the result as acceleration and calculated the time and distance from it.
It was movable and never hit limits.
I never twigged but might have tomorrow when the machine pulled itself off the floor!

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equation to find minimum acceleration

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